Functions of a Complex VariableDate: 10/05/2002 at 11:28:34 From: Jay Subject: Functions of a complex variable Show that w=f(z*), where z*=x-iy and f is a differentiable function, is not an analytic function of z. Date: 11/11/2002 at 11:43:14 From: Doctor Nitrogen Subject: Re: Functions of a complex variable Hi, Jay: This gets pretty involved, so I hope you have time to read carefully. The only proofs I could find somewhat related to my argument here can be found on page 71, problem 2, and page 32, problem 5,in: _Theory and Problems of Complex Variables_, Murray Spiegel, McGraw Hill, 1964, Schaum's Outlines, ISBN #07-060230-1. Do you recall those delta(y)/delta(x) equations you had to use when you first learned to calculate a derivative in calculus? That is, if you recall, you had to compute a limit of something like [f(x + delta(x)) - f(x)]/delta(x), to find the derivative of something like f(x) = x^2. To prove that f(z*) fails to be analytic, you will have to do something similar. First of all, let w = f(z*) = ((u(x, y) + iv(x, y))*. By the definition of derivative in Complex Variable Theory: (d/dz*)f(z*) = Limit [f((z + delta(z))*) - f(z*)]/(delta(z*)). delta(z*)->0 First, consider what happens when delta(y) = 0, lim delta(x)-> 0: When (delta(y) = 0), Limit [f((z + delta(z))*) - f(z*)]/(delta(z*)) delta(x)->0 = Limit { [u(x + delta(x), y) - iv(x + delta(x), y)] delta(x)->0 - [u(x, y) - iv(x, y)] } / delta(x) = Limit { u(x + delta(x), y) - u(x, y) delta(x)->0 - i[v(x + delta(x), y) + iv(x, y)] } /delta(x) = Limit { u(x + delta(x), y) - u(x, y) } / delta(x) delta(x)->0 - [i(v(x + delta(x), y) - iv(x, y)] / delta(x) [A] = &u/&x - i&v/&x, where "&this/&that" denotes "the partial derivative of this with respect to that." Now consider what happens when delta(x) = 0 and lim delta(y) -> 0: When (delta(x) = 0), Limit [f(z + delta(z))*) - f(z*)]/(-i delta(y)) delta(y)->0 = Limit { [u(x, y + delta(y)) - iv(x, y + delta(y))] delta(y_->0 - [u(x, y) - iv(x, y)] } / (-i delta(y)) = Limit { [u(x, y + delta(y)) - u(x, y)] delta(y)->0 - [iv(x, y + delta(y) + iv(x, y)]} / (-i delta(y)) = Limit { [u(x, y + delta(y)) - u(x, y)] delta(y) + [iv(x, y + delta(y) - iv(x, y)] } / (i delta(y)) [B] = &u/&y + i&v/&y. Now recall that for a complex valued function to be analytic, a necessary and sufficient condition is for the Cauchy-Riemann equations to be satisfied, that is: [C] &u/&x + i&v/&x = &v/&y - i&u/&y, or &u/&x = &v/&y, &u/&y = -&v/&x. But compare what you must have for f(z*) to be analytic from [A] and [B]: [A] &u/&x - i&v/&x [B] &u/&y + i&v/&y, or it must be that [D] &u/&x = &u/&y, &v/&y = -&v/&x. Clearly, [C] and [D] are not the same conditions, are they? But it is known that [C], or the Cauchy-Riemann equations, must be satisfied, so [C] is true, meaning the equations in [D] cannot be true, so f(z*) can't be analytic. You also might ask yourself, as another way to prove f(z*) fails to be analytic, (I did not investigate this myself, but you might want to) whether both Re[f(z*)] and Im[f(z*)] are harmonic functions, since every analytic function has two harmonic functions associated with it, that is, f(z) = u(x, y) + iv(x, y) is an analytic function provided u(x, y) and v(x, y) are both harmonic. If you can show Re[f(z*)] and Im[f(z*)] are not harmonic, it would follow they are not analytic. I hope this helped answer the questions you had concerning your mathematics problem. You are welcome to return to The Math Forum and Doctor Math whenever you have any math-related questions. - Doctor Nitrogen, The Math Forum http://mathforum.org/dr.math/ |
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